1

Engineering Wallah Electrical Engineering

SHORT NOTES
Control Systems
CHAPTER
Basics of Control System
1

Shifting
(a) f(t + 1) → advance
(b) f(t – 1) → delay

(a) f(t + 1), t = –1 (b) f (t – 1), t = 1

f (t  1)  2 ;  2  t  0 f (t  1)  2 ; 0  t  2
0 ; otherwise 0 ; otherwise

Scalling
(a) Compression f(2t)
(b) Expension f(t/2)

(a) Compression f(2t) (b) Expension f(t/2)

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Time domain representation:

Shifting & scaling:
Amplitude not changes.
Reversal

If all three properties will be asked in ques together, then

Shifting → Scaling → Reversal
Some Basic Signals
Impulse signal
(I) (t) = 0, t  0
(II) (t)  0, t = 0

(III)  (t )dt  1


Example:

1
(I) Width 
height

1
(II) (II)Area of impulse = 1 ,  (t )dt  1, Area  height  width  a  a 1

 3

Shifting:

(a) (t – 1), (t) = (–t) (even signal)

(b) (t + 1), (t) = – (–t) (odd signal)

(a)  (t – 1) (b)  (t + 1)

Scaling:

1 1
(at )   (t ) e.g :  (2t )   (t )
|a| 2

Multiplication Property:
f (t )(t  a)  f (a). (t  a)

t+a=0
t=–a

Integration Property:

 f (t ).t (t0) at  f (0)
 
 f (t ).t (t1 1)0 dt  f (1)  f (t ).t(t 202) dt  f (2)
t 1 t 2

Step Signal
Standard form, u(t) = 1, t > 0
0, t < 0
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a). u(t – 1) = ? b). u(t + 1) = ?

Ramp Signal

Standard form. r(t) = t, t ≥ 0

0, t < 0

Exponential signal

Oscillatory Signal

Parabolic signal
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Damped Oscillatory

Laplace Transform

It is a mathematical tool, used to solve linear differential equations. It converts a differential equation. Into an algebraic
equation in s-domain.

Unilateral Laplace Transform

LT
(I) (t ) 
1 (II). 1.u (t ) 
1/ s

1.(t ) 
1 A.ut 
A/ s

A. (t ) 
A

1
(III) e  at u (t ) 

sa
1
eat u (t ) 

sa
Properties of Laplace Transform
1. Shifting property
1
f (t ) 
 F ( s) u (t ) 

s
1
 F ( s )e  sto u (t  2) 
f (t  to )   e2s
s
1
 F ( s)e sto
f (t  to )  u (t  2)  e 2 s
s
2. Linearity Property
 6

af1 (t ) 
 aF1 (s)

bf 2 (t ) 
 bf 2 ( s)

af1 (t )  bf 2 (t ) 
 af1 (s)  bF2 (s)

3. Scaling Property

f (t ) 
 F ( s)

1
f (at ) 
 F ( s / a)
|a|

4. Differential Property

d
f (t ) 
 F (s) f (t ) 
 sF ( s )  f (0)
dt

d2 d F ( s )
f (t ) 
 s 2 F (s)  sf (0)  f '(0) t. f (t ) 

dt ds

Example:

1 d  1  1
u (t ) 
 t.u(t ) 
  
s ds  s  s 2

d  1  (2) 2 n!
t 2u(t ) 
   3  3 t nu (t ) 

ds  s  s s s n1

LT 0 LT s
sin 0t u (t ) 
 cos 0t u (t ) 

s 2
 02 s  02
2

5. Time Integration Property

t 1
0 f (t )dt 
LT LT
f (t ) 
 F ( s)  F (s)
s

Initial Value Theorem

LT
f (t ) 
 F ( s) Lt f (t ) 
 f (0)  Lt sF (s)
t 0 s

Example:

1 ( zeros)
f ( s) 
s  1 ( poles)

 1   s  s 1 1
Lt sF (s)  Lt s    Lt    Lt   1
s 0 s   s  1  s  s  1  s   1 1
1 1 0
s 1  
 s 

Initial value theorem is applicable only in proper form function.

Proper form → order of a denominator should be less than order of numerator , order → No. of poles of the system.
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Final Value Theorem

LT
f (t ) 
 F ( s)

LT
Lt f (t )  f () 
 Lt sF (s)
t  s

Example:
1
F (s) 
s 1

 1   s  0
Lt sF (s)  Lt s    Lt   0
s0 s0  s  1  s0  s  1  0  1

Find value theorem is only applicable for stable system.

Improper form → Order of denominator should be greater than order of numerator.

(t) → impulse function r(t) → ramp function

u(t) → step function p(t) → parabolic function
Transfer Function:

It is the ratio of LT of output to the LT of input

C(s) = R(s). G(s)
C (s) output
G( s)  
R(s) input
C (s) (s  z1 )(s  z2 ).......(s  zn )
G( s)  
R(s) (s  p1 )(s  p2 )....(s  pn )

s  z1  0  s   z1

Zeros →  z1,  z2 ,  z3.....,  zn

Poles →  p1,  p2 ,  p3.....,  pn

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sz
G( s) 
s ( s ' p1 )( s ' p2 )
k

k = number of poles at origin = type

Order = (k + 1 + 1) = k + 2
Example:
s 1
, type→ 4, order → 6
s ( s  2)( s  1)
4

Example:
s 1
G(s)  , type→ 5, order → 8
s ( s  3)( s  1)(s  2)
5

Control System:
Gives desired output with the help of an arrangement of physical components.

Ex→Fan
1. Open loop control system (LCS) manually controlled

It is the control system in which controller action does not depend on the output.
Example:
Washing machine
Toaster
Indian Traffic System etc.
2. Closed loop Controlled System
It is the control system in which controller action depends on the output.

Example:
Automatic air conditioner (AC)
Automatic Gyser
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Human Beings
Transfer function is valid only for linear time invariant (LTI) systems.
1
B.W.  ,  time constant

S =  + jw
1 1
Time constant ()  
Real part 

Transfer function for OPCS:

C ( s)
 , Initial condition = 0
R( s )

Transfer function for CLCS

C ( s) G(s)

R( s ) 1  G ( s ) H ( s )

C ( s) G( s)
Negative feedback → TF  
R( s ) 1  G ( s ) H ( s )

G( s)
Positive feedback → TF 
1  G( s) H ( s)

Positive feedback is used in oscillator & multivibrator.

G( s)
Gain  T (s) 
1  G( s) H ( s)

Positive feedback → Regenerative feedback

Negative feedback → Degenerative feedback
Negative feedback is used in amplifier circuit
Negative feedback reduces the overall gain.
Positive feedback increases the overall gain.
Negative feedback increases → stability
Positive feedback reduces → stability
Negative feedback reduces the time constant (). And increases bandwidth.
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1
BW 

Sensitivity: It is the measure of change in output due to change in input or its internal parameter
Requirement for good control system.
System should be highly sensitive to the input.
System should be very less sensitive to the internal parameter.
How to find sensitivity mathematically
A / A
A  f ( B) S BA 
   B / B
change change sensitivity of A w.r. to B.

B A
S BA 
A B
Let, A1 → initially, A2 → final
Change = A2 – A1 = A
A
% change in A   100%
A
B
Lly, % change in B   100%
B

Open loop control system (OLCS): SGT  1

1
Closed loop control system (CLCS): SGT   1 (less than 1)
1  GH
1 + GH > 1
System modeling:
1. Electrical system
2. Mechanical system

Force voltage
Translation mechanical Rotational Mechanical Force current analogy
analogy (Electrical
system (TMS) System (RMS) (Electrical system)
system)
1 1 1
F = kx + Bx + Mx Τ = k + B + j V q  Rq ' Lq " I    ' C  "
C E R
F Τ V I
Force Torque Voltage Current
M J L C
Mass Moment of inertia Inductance C→capacitance
x  q 
linear displacement Angular displacement Charge Magnetic flux
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B B R 1/R
Damping coefficient Damping coefficient Resistance R→Resistance
K K 1/C 1/L
Spring constant Spring constant C→ capacitance L→inductance

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CHAPTER
Block Diagram & Signal Flow Graph
2
Block diagram Reduction
It is the pictorial representation of control system.

G( s)
TF 
1  G( s) H ( s)

Block Reduction Technique

C ( s) G( s)
1. 
R( s ) 1  G ( s ) H ( s )

Summing point.

2.

Series = multiply,
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3.

Parallel = addition

4. Shifting of take-off point before a block

5. After a block

Before → G(s), add

1
After → , add
G( s)
6. Shifting of take-off point after the summing point.

7. Before summing point

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8. Shifting of Adder/summing point before a block

9. After a block]

Before → 1/G, After → G

Q :- Find the T.F of the system

C A( B  C )

R 1   A( B  C )

C AB  AC

R 1  AB  AC
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Signal flow Graph

Graphical representation of control system.

Input node → having only outgoing branches. (x0).

Output node → having only incoming branches (x7).
Chain node → having both incoming and outgoing branches. (x1, x2, x3, x4, x5, x6)
Forward path → Path connected between input to output.
To find forward path, node should not be repeated.
Forward path = 2(P1 and P2)
P1 = x0 → x1 → x2 → x3 → x4 → x5 → x6 → x7
P2 = x0 → x1 → x2 → x3 → x5 → x6 → x7
Forward path gain → gain of,
P1 → 1  a  b  e  f  i  1
P1 → 1  a  b  g  i  1
6. Feedback loops →

L1 →

L1 = x 1 – x 2 – x 3 – x 1
gain → L1 = abc

L2 →
 16

L2 = x 2 – x 3 – x 4 – x 2

gain → L2 = beh

L3 →

L3 = x 5 – x 6 – x 5

gain → L3 = i × 1 = i

7. Self loop →

L4 = x 3

gain = d

8. Non – touching loop →

L3 = x 5 – x 6 – x 5

L1 and L2 → touching loops

L1 and L3 → non touching loops

L2 and L3 → non touching loops

L2 and L4 → touching loops

L3 and L4 → non touching loops

L2 and L3 → non touching loops

Mason’s Gain Formula :-

It is used to calculate the T.F of any SFG of given system.

C ( s) n Pk  k
Transfer function (T .F .)  
R( s) k 1 

K = 1, 2, 3, ......... , n

C (s) P11  P2 2  .........Pn n


R( s ) 
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Where,

P1 = Forward path gain of path ‘1’

Pn = Forward path gain of path ‘n’

1 Calculation :-

Step :- Remove the forwad path 1, Remove all the nodes that comes in FP – 1.

Step :- 1 – (L1 + L2 + L3) + (L1L2 + .....)

2 Calculation :-

Same as 1 calculation and so on.

Limitations of Mason’s gain formula :-

1.

If self loop of gain = 1 exist at any node then that loop is invalid

P11 a1a2 1 a1a2

T F   
 1 – a3 1 – a3

If a3 = 1 then,

2.

→ If input node has a self loop of any gain then that will be invalid loop.

→ Mason gain formula not valid.

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

CHAPTER
Time Domain Analysis & Study State Error
3

Time Domain Analysis

Time Response

Steady State : y(t) output does not change with respect to time.

Transient : y(t) output changes with respect to time

1
(I) Stability 
transient time

1
(II) Relative stability 
transient time

1
(III) Speed of response 
transient time

Transient ↑, stability ↓, speed ↓

1st Order System : (1 pole)

1
G (s)  , where τ = time constant.
s '

1
Time constant 
real part of pole
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Question:

C ( s) 1
 , partial fraction → inverse LT
R( s ) 1  s

 
c(t )  1  et /  u (t )

 
t  0, c(0)  1  e0/  u (t )  0

 
t  1, c()  1  e /  .u (t )  1

τ↑, 4τ↑, speed ↓, stability ↓, pole close to origin ↑

(and vice-versa).....

DC Gain :

Frequency = 0, s = 0

Its order system

1
G(s)  , impulse response
1  s

Pole = 1

Type = 1

DC gain = 1

Standard form of 1st order system

k
TF  ,where k = DC gain
1  s

k
DC gain  Lt sn , n→ type of the system
s 0 s (1  s)
n

2nd Order System : (2 poles)

n 2
OLTF = G ( s ) 
s 2  2n s

G( s)
CLTF 
1  G ( s) H ( s)

H(s) = 1,

n 2
CLTF  , Standard form
s 2  2n s  n2
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k .n 2
CLTF 
s 2  2n s  n 2

Where, k = DC gain

 = Damping factor

n = Undamped natural frequency

Second order system characteristic equation

1  G ( s) H ( s)  0

s 2  2n s  n2  0

Stability depends on  (zita)

C(t) = Impulse
Damping Pole-Zero diagram Roots Stability, τ, FOD
Response

Imaginary
(I) Marginal stable
1. =0 roots lie on
(II) τ = ∞
undamped the
(III) FOD = n rad/sec.
system imaginary
FOD – frequency of damping
axis

Complex (I) Stable

2. 0<<1 1 1
conjugate (II)   
Under
damped
poles lies n (alpha)
on the LHS
system
of s-plane. (III) FOD = d rad/sec.

3. =1 (I) Stable

Two real 1
critical (II)   sec.
negative
damped
equal poles
n
system (III) FOD = 0 rad/sec.

(I) Stable
4.  > 1 over Real (II)  
1
damped unequal n  n  2  1
system poles
(III) FOD = 0 rad/sec.
Slow system, less stable
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Damping c(t) = step response Pole-zero diagram Roots Stability, , FOD

Conjugate
Imaginary
1. =0 (I) Margin stable
poles lies
Undamped (II)  = ∞
on the
system (III) FOD = n rad/sec.
imaginary
axis

Comples (I) Stable

2. 0<<1
conjugate 1 1
Under (II)   
poles lies
damped n 
on the LHS
of s-plane (III) FOD = d rad/sec.
system

3. =1 (I) Stable

Real 1
Critical (II)   sec.
negative
damped
equal roots n
system (III) FOD = 0 rad/sec.

(I) Stable
4. >1 Real
Over negative 1
(II)  
damped unequal n  n  2 
system roots
(III) FOD = 0 rad/sec.

The domain parameters:

Peak time:
Time during which response either maximum or minimum.
n
tp  , where, d = damping frequency, d  n 1  2
d

n = 1, 2, 3, 4,......
 22

(2m  1)
n → odd, overshoot time t p0 
d

n → even, undershoot time t pu 

 2m   , m = 1, 2, 3,......
d

Maximum Peak Overshoot (MPO):

c(t p )  final value
MPO  100%
final vlaue

12
% MPO  e / 100%

Undamped ( = 0) = % MPO = 100%

Critically damped ( = 1) = % MPO = 0%

2. Settling Time (ts) :
Control system → 2% to 5% error can be accepted.
3. Rise Time:
(i) Underdamped system:


0% to 100% , tr  ,   angle of inclination
d

(ii) Critically damped system:

5% to 95%
(iii) Overdamped system:
10% to 90%
4. Delay time (td):
Time taken to reach 50% of its final value.

ts  t p  tr  td
 23

Steady state error ess (t ) :

When the system attained the steady state, then the difference between the desired value and the system response at
steady is knows as steady state error.
Lt ess (t )  Lt s E ( s )
t  s 0

R( s )
E ( s) 
1  G( s)

R( s )
Lt ess (t )  Lt s.
t  s 0 s  G( s)

Stead state error for various inputs.

Input Steady state error

1
Step ess  k p  Lt G ( s )
1 kp s 0

1 kv  Lt s G ( s )
Ramp ess 
kv s 0

1
Parabolic ess  ka  Lt s 2 .G(s)
ka s 0

Steady state error comparing in terms of OLTF:

Type kp kv ka u(t) t.u(t) T1/2.ut
k k
0 0 0 ∞ ∞
(finite) (finite)
k k
1 ∞ 0 0 ∞
(finite) (finite)
k k
2 ∞ ∞ 0 0
(finite) (finite)

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

CHAPTER
Routh Hurwitz {R–H} Criterion
4

System stability: → decide → poles of close loop TF.

C s G s
CLTF T .F  
Rs 1 G s H s

For unity feedback, H(s) = 1.

Characteristics equation (CE)

1 + G (s) = 0 → Roots → poles of close loop TF → decide system stability.

Note: G (s) = OLTF → stable

But it does’t mean that CLTF also be stable.

Poles at origin → stable

Repeated poles at origin → unstable

Pole at imaginary axis → stable

Repeated poles at imaginary axis → unstable

R-H criteria:
• Used to find stability of the system.
• Determines the no. of poles on LHS, RHS & on the imaginary axis of s-plane.
• Frequency of oscillation & DC gain can be calculated.

How to find characteristic Eqn (CE)

10
G s  , H s  2
 s  1 s  2 
CE 1  G  s  H  s   0

10
1  20
 s  1 s  2
 s  1 s  2  20  0
s 2  3s  2  20  0

s 2  3s  22  0
 25

LHS = Left half of s-plane

RHS = Right half of s-plane

How to make Routh table

a0 s m  a1 s m1  a2 s m2 ............. an s mn  0

Ex:
(i) s4  s3  s 2  2s  1  0

(ii) s3  2s 2  4s  6  0
(iii) s 2  3s  22  0

Two things to remember:

1.

Unstable System

2. no sign change but missing a term.

Unstable system

Ques:- 2s4  s3  3s2  5s  10  0

(i) How many poles lies on LHS.
(ii) How many poles lies on RHS.

Sol:
Make Routh Table.
Divide

1 3  2  5
+ S4 2 3 10 a1   7
1

1  10  0  2
+ S3 1 5 0 b2   10
1

1 0  0 1
– S2 -7 10 0 c1  0
1

45 7  5  10  1 45
+ S1 0 0 a2  
7 7 7
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7  0  0  1
+ S0 0 0 0 b2  0
7

Note:- 1st column esa ftruh ckj sign change gksxk] mruh gh poles RHS djsx
a sA
2 times sign change means 2 roots lies on Right half of s-plane.
Total poles = 4

(i) 2 poles lies on RHS.

(ii) 4 – 2 = 2 poles lies on LHS.

Unstable System
Special Cases
Case-I
When any element of the first column will become zero & in that row at least one non zero element will present.

Ex:

S 5  S 4  2S 3  2S 2  3S  5  0

RH Table

22 35 00

S5 1 2 3 s3   0,   2, 0
1 1 1
S4 1 2 5 Replacing 0 by ∈ (eplsilon)

2  2 5  0
S3 0/∈ –2 0 s2  , ,0
 
 2  2  4  4
2   5
2  2  
S2 5 0 s1    
 2  2 2  2
 
4  4  5 2
S1 0 0
2  2
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4  4  5 2 4  4  5 2

S0 5 0 0  2  2
2  2



∈ = avg positive no.

First column → + + + + – +
Sign change = 2 = 2 poles lies on RHS
Total poles = 5
LHS = 5 – 2 = 3 poles

Case-II
When any odd row of Routh table becomes zero.
Ex:
s6  2s5  8s 4 12s3  20 s 2 165  16  0
S6 1 8 20 16
5
S 2 12 16 0
4
S 2 12 16 0
3
S 0 0 0 0 → all zero
S2
S1
S0
Now we will make auxiliary eqn
A  s   2s 4 12s 2  16  0 _________ (i)

d A  s   8s 3  24 s  0
ds  8s 3  24 s
Now,
S3 8 24 0 0
2
S 6 16 0 0
1
S 8/3 0 0 0
S0 16 0 0 0

No sign change = RHS → poles = 0

From eqn (i), s2 = k

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2s4  12s2 16  0

2k 2  12k  16  0
k 2  6k  8  0
k 2  4k  2k  8  0
k  k  4  2  k  4  0

 k  2 k  4  0
k   2,  4
Imaginary axis is lie djsxk

Case-III
When variable is associated in the characteristic eqn.
Ex: s3  2s 2  5s  k  0
Sol:
S3 1 5
2
S 2 k
S 1
10  k
0
2
S0 k

Q1. If system will be stable then value of k ?

Sol.
stable, k > 0
10  k
0
2
10 – k > 0
k > –10
0  k 10

Q2. If system is unstable then value of k ?

Sol.
Unstable, k > 10

2 times sign change → unstable system

Q3. If system is marginal stable then value of k ?

Sol.
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Marginal stable
10  k
0 s2/2 10
2
10 – k = 0 2s2 + 10 = 0
k = 10 s2 + 5 = 0
s2 = –5
s = ± j5

Short trick for 3rd order system

a s3  b s2  c s  d  0
External product Internal product
(EP) = ad  IP   bc

(i) EP > IP (ad > bc) → Unstable

(ii) EP < IP (ad < bc) → Stable

(iii) EP = IP (ad = bc)→ marginal stable

Ex:

s3  2s 2  5s  k  0 , Stable, IP > EP ⇒ 10 > k , (∵ 0 < k < 10)

Advantages of Routh's criterion:

 We get the stability of the system easily, without solving the characteristic equation.
 Relative stability of the system can be easily judged.
 Critical value of system gains, frequency of sustained oscillations can be determined.
 Inter section points of root locus with imaginary axis can be find out.

Limitation of Routh:

 Applicable to finite order polynomial only.

 It is not valid for complex coefficient.
 It only provides the information about number of roots on right half of s-plane.
 It cannot distinguish between the real and complex roots.


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